Chapter 19: Q61P (page 580)

A gas is to be expanded from initial state i to final state f along either path 1or path 2on a PV diagram. Path1 consists of three steps: an isothermal expansion (work is $40 J$ in magnitude), an adiabatic expansion (work isin magnitude), and another isothermal expansion (work is $20 J$ in magnitude). Path2 consists of two steps: a pressure reduction at constant volume and an expansion at constant pressure. What is the change in the internal energy of the gas along path 2?

Expert verified

The change in internal energy along path 2 is $- 20 J$ .

Step by step solution

The expression for the heat from the first law of thermodynamics is given by,

$Q = W + Δ U$

Here Q is the heat required, W is the work done, $Δ U$ is the change in internal energy.

So, the first law of thermodynamics gives the relation between heat, work and internal energy.

For path 1:

As internal energy is a state function for an isothermal process, $d U = 0$ .

For adiabatic process, $d Q = 0$ ,

Therefore, $d U = - d W$

Therefore, net change in internal energy of path 1 is

$d U = - 20 J$

But, internal energy is independent of the path, so change in internal energy along path 2 is the same as path 1.

$d U = - 20 J$

Therefore the change in internal energy along path 2 is $- 20 J$ .

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